Re: Non-Scalar Cryptography - The Emporor is stark naked.

On May 15, 11:07 pm, rossum <rossu...@xxxxxxxxxxxx> wrote:
On Wed, 14 May 2008 11:03:14 -0700 (PDT), AdaCrypt

<austein.oby...@xxxxxxxxxxxxxx> wrote:
The algorithm is:

(Plaintext +key) Mod 95 = Ciphertext Mod 95

The second mod 95 is redundant:
  (plaintext + keystream) mod 95 = ciphertext



Analysing the key probability as you have done will not affect the
uncertainty of the ciphertext => the ciphertext still has equal
probability between elements = > the ciphertext string is random by
definition surely => this is the backbone of this cipher - whether it
is an OTP or not is not important ?? - I am taking on board your
suggestion regarding a standalone cipher - Thanks for your help

The key *size* is important.  To take an extreme example, assume that
the scrambling key is an absurdly small 16 bits; that means that there
are only 65536 possible different values of the key, so there are only
65536 possible different permutations that your scrambling algorithm
can generate from that size key.  An attacker can brute-force that
number of permutations reasonably quickly.  There are many potential
permutations, but you are only able to use the permutations generated
by your scrambling key.  The bigger the key, the bigger the number of
possible permutations, up to about 2800 bits.  After that point you
will have key pairs which generate the same permutation from different
keys.

The permutation generated by your scrambling key does not have equal
probability between the elements since the probability of an element
appearing is dependant on the number of times that element has already
appeared in the permuted list of 14,000 characters.  This is a
potential source of weakness.

rossum

Thanks again. You are considering that an attack of some sort has
given the attacker access to the current key string ( that is normally
stored away safely by Alice) and the adversary now has both the
ciphertext string and and the corresponding key string to hand. It is
then a question of identifying the key by probability methods. I
appreciate that there is a scene here for an attack. I don't think
however that there is feasible computation over the entire ciphertext
string by this means although I admit it is an uneasy state. Remember
also that this will not compromise the entire scheme because Alice is
about to change it all again in her next message.

Your analysis implies to me that although the cipheretxt string is
designed by me to be random it is now not random because unequal
probability is being injected by the probability of the key being X
say in the algorithm,

PlainText Mod 95 + Key Mod 95 = Ciphertext Mod 95

How certain can you say your knowledge of the key would be at the end
of your attack that is based on the probability of this key being X
say.. If there is any remaining uncertainty ( less than 100%
probability) about the key, even if the message you decipher appears
to make perfect sense, then that uncertainty goes across by default
to the corresponding ciphertext. Since your method is the same for
all items then then the equal uncertainty over the entire string
remains unchanged ( since the argument is the same for each item )
and the ciphertext string is still random albeit at some different
overall level of absolute (unquantifiable) uncertainty. I contend
that the ciphertext is still random and thus remains secure ! - Thanks
for your great help - Adacrypt.
.

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